
#19
Mar912, 06:37 PM

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PF Gold
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There are always an infinite number of ways to choose a complementary subspace. The simplest example is in R². Set V:=R x {0}. There is an infinite number of ways to chose a subspace complementary to V since in fact, any subspace other than V itelf does the job, so the complementary subspaces H are in 11 correspondence with the angle θ in (0,pi) that they make with the horizontal.
This illustrates the situation in the linear setting. In the vector bundle setting there is even more freedom in the choice of a complementary subbundle H. 



#20
Mar912, 06:41 PM

P: 2,053

Ah, that makes things a lot clearer. Thanks.
EDIT: Just out of curiosity, are you a mathematician? Or a physicist who learned a lot of math? hehe... 



#21
Mar1012, 05:21 PM

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Horizontal vectors are tangent to the bundle of frames. They vary with the connection and are not in the kernel of the projection. 



#22
Mar1012, 05:39 PM

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#23
Mar1012, 05:44 PM

P: 2,053

The book I'm using defines parallel transport along a curve c(t) in M of a vector in an associated vector bundle (taken to be TM for simplicity) as follows. Given the curve c(t) in M, horizontally lift this curve into the frame bundle giving you a curve c'(t) in the frame bundle such that c'(0)=b, where projb=c(0). This curve c'(t) gives you the parallel translate of a the point b along the curve c(t), and gives you, at each point, the parallel translated basis vectors for our associated tangent bundle {f(t)}. The parallel transport of a vector in the tangent bundle along this curve c(t) is then basically to keep the components of our vector constant as we change our basis vectors so that they are parallel transported. I.e. v(t)=sum(v(0)f(t)) where v(t) is the parallel transported vector, and v(0) are the components of this vector as t=0.
This construction makes sense to me. In essence, I am just parallel transporting my basis vectors using this connexion H on the frame bundle, and that gives me a notion of parallel transporting my vectors in TM by keeping their components constant since my basis vectors are parallel. This means that my connexion H on my frame bundle induces a connexion H' (equiv to a law of parallel transport) on my tangent bundle. My question is, is there some (not ridiculously difficult) way to link this "new" construction of parallel transport, etc., with the "regular" construction I'm used to in the sense of a covariant derivative? I.e. is there some way to pinpoint which H I should choose, if I have to choose, i.e. on a Riemannian manifold the unique law of parallel transport that preserves inner products and has no torsion? This should pinpoint for me a unique distribution H in my frame bundle right? But how do I know which H? EDIT: I have an undergrad degree in Astrophysics and Economics, and am a graduate student in physics. I'm learning this mostly for my own enjoyment because I have a lot of interest in differential geometry, but also because I want a clear mathematical understanding of the concepts I use in general relativity and quantum field theory for connections, etc. =] 



#24
Mar1012, 06:10 PM

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You can look at Chris Wendl's remarkably enlightening lecture notes on bundles (see his website). Particularly, Theorem 3.38 exhibits the link between connexions on the frame bundle and the corresponding connexions in the associated vector bundle.
here:http://www.homepages.ucl.ac.uk/~ucah...nnections.html 



#25
Mar1012, 06:44 PM

P: 2,053

Ok, I'll take a look thanks =]




#26
Mar1012, 06:56 PM

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An Example: The torus is a circle bundle over the circle and has a natural action of SO(2) on each fiber  just rotate. Any vector field that is invariant under fiber rotation and which is never tangent to any fiber circle generates a horizontal space. The connection one form is easy to define from the decomposition of a vector into horizontal and verticle components.
to see that there are infinitely many connections, just take any smooth vector field on a circle that is transverse to the fiber and rotate is around to get a vector field on the whole torus. 



#27
Mar1012, 08:13 PM

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An example: (Please correct this if it is wrong.)
On a surface (two dimensional Riemannian manifold) Take a curve that is parameterized by arc length and a unit vector field ,v,along the curve that is perpendiclular to the curve. The covariant derivative of v will be tangent to the curve and so is some multiple of c'(s). To compute this multiple view v as a map of the curve into the tangent circle bundle (same as the bundle of orthonormal frames). dv(c'(s)) is the derivative of v with respect to c'(s). Decompose dv(c'(s)) into horizontal and vertical components. The multiple we are looking for is the coefficient of the vertical component of this decomposition. c'(s )is also a map of the curve into the tangent circle bundle and the argument is the same. The covariant derivative this time will be in the v direction normal to c'(s). Extend now by the Leibniz rule and linearity to arbitrary v. 



#28
Mar1312, 12:25 AM

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When you refer to the dimension of Gl(4,R) ; algebraic dimension, if I understood well, is equal to 4^2=16. What kind of algebraic structure are you assigning to Gl(4,R)? 



#29
Mar1312, 12:32 AM

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GL(n,R) is an open subset of ##M_n(\mathbb R) \cong \mathbb R^{n^2}##, so it's a manifold of dimension equal to dim M_n(R) = n^2.




#30
Mar1312, 12:44 AM

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I understand that, but I thought I read a reference to algebraic dimension, tho I may have misunderstoodor misunderestimated what I read.




#31
Mar1312, 07:46 AM

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No reference to algebraic dimension. :)




#32
Mar1312, 09:40 PM

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Sorry, Quasar,
I misunderestimated your question :). Nice post, btw. 



#33
Mar2212, 09:04 PM

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Sorry again for my slowness, Quasar. I understand that Gl(n,R)=Det^{1}(ℝ\{0}), which is open, yada, yada; I realized where my confusion lay.
Anyway, another dumb question; I am also trying to teach myself some bundles: (my apologies, my quoting button is disabled for some reason): say you are considering the case of a vector bundle ( thinking mostly of tangent bundle), in a situation where you have a metric defined. It would then make sense to define the complement bundle to be the orthocomplement (with respect to this metric), right? Is there anything special to this choice? 



#34
Mar2312, 07:51 AM

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I'm not sure I follow you. You are thinking of the case where (M,g) is a riemannian manifold. Then you have the projection map pr:TM>M and its differential pr_{*}:T(TM)>TM and you want a complementary subbundle to V:=ker(pr_{*}). You cannot just pullback g to T(TM) via pr, and use that to define H:=V[itex]^{\perp}[/itex] because pr^{*}g is very degenerate: pr_{*} vanishes on V and so V[itex]^{\perp}[/itex]=T(TM).
Is this what you were thinking? 



#35
Mar2312, 04:40 PM

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Yes, got it, thanks. I just need to read more carefully. Sorry.




#36
Mar2312, 06:16 PM

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For the sake of background: I ended up studying bundles by mistake:
My girlfriend wanted to take a class in cosmetology, but she misread the instructions in the webpage, and ended up registering for cosmology instead . Since there were no refunds, and she knew no math,I had to help her. Then I became interested in bundles. 


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